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If `.^(n)C_(8)=.^(n)C_(6)`, then find `.^(n)C_(2)`. |
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Answer» If `.^(n)C_(x)= .^(n)C_(y) " and" x ne y`, then x+y=n. Hence, `.^(n)C_(8)= .^(n)C_(6)` `implies n=(8+6)=14` Now, `.^(n)C_(2)= .^(14)C_(2)=(14xx13)/(2)=91` |
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